A077657 Least number with exactly n consecutive successors, all having the same number of prime factors (counted with multiplicity).
1, 2, 33, 603, 602, 2522, 211673, 3405123, 3405122, 49799889, 202536181, 3195380868, 5208143601, 85843948321, 97524222465
Offset: 0
A077655 Number of consecutive successors of n having the same number of prime factors as n (counted with multiplicity).
0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0
Offset: 1
Keywords
Comments
If a(n) > 0 then a(n+1) = a(n)-1.
Examples
33=3*11 has only two successors also with two factors: 34=2*17 and 35=5*7 (whereas 33+3=36=2*2*3*3), therefore a(33)=2.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..100000
Programs
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Mathematica
snpf[n_]:=Module[{f=PrimeOmega[n],k=0},While[f==PrimeOmega[n+k],k++];k]; Array[snpf,110]-1 (* Harvey P. Dale, Aug 01 2021 *) -
PARI
A077655(n) = { my(k=n+1,w=bigomega(n)); while(bigomega(k)==w,k++); (k-n)-1; }; \\ Antti Karttunen, Jan 22 2020
Comments
Examples
Crossrefs
Formula
Extensions